我想从python中提取zip文件中的特定文件夹,然后在原始文件名后重命名。
例如,我有一个名为test.zip
的文件,其中包含多个文件夹和子文件夹:
xl/media/image1.png
xl/drawings/stuff.png
stuff/otherstuff.png
我希望将提取的媒体文件夹的内容提供给名为test的文件夹:
test/image1.png
答案 0 :(得分:7)
使用
zipfile
模块,尤其是ZipFile.extractall()
os.path.splitext()
从字符串test1
test1.zip
tmpfile.mkdtemp()
创建临时目录shutil.move()
移动整个目录树。例如:
#!/usr/bin/env python
"""Usage:
./extract.py test.zip
"""
from zipfile import ZipFile
import os
import sys
import tempfile
import shutil
ROOT_PATH = 'xl/media/'
zip_name = sys.argv[1]
zip_path = os.path.abspath(zip_name)
extraction_dir = os.path.join(os.getcwd(), os.path.splitext(zip_name)[0])
temp_dir = tempfile.mkdtemp()
with ZipFile(zip_path, 'r') as zip_file:
# Build a list of only the members below ROOT_PATH
members = zip_file.namelist()
members_to_extract = [m for m in members if m.startswith(ROOT_PATH)]
# Extract only those members to the temp directory
zip_file.extractall(temp_dir, members_to_extract)
# Move the extracted ROOT_PATH directory to its final location
shutil.move(os.path.join(temp_dir, ROOT_PATH), extraction_dir)
# Uncomment if you want to delete the original zip file
# os.remove(zip_path)
print "Sucessfully extracted '%s' to '%s'" % (zip_path, extraction_dir)
使用try..except
块来处理创建目录,删除文件和解压缩zip时可能发生的各种异常。